Monday, June 27, 2016

Any equation having
in the form of ax2+bx+c=0,is a quadratic
equation in where x represents an unknown. a, b &
c represents the real numbers such that a is not equal
to 0.

In this Equation:

a is the coefficient of x2

b is the coefficient of x

c is the constant term

How
to solve Single Variable Quadratic Equation ?

In single variable quadratic equation
we will be given one quadratic equation in the form of ax2+bx+c=0 & we have
to find the value of that unknown variable x.

Let us take an Example:

Example: Let we have
given the equation: 4 x2+12x+8=0

Solution:

In the above given equation, we have

+4 is the coefficient of x2

+12 is the coefficient of x

+8 is a constant value

Step-1: Multiply the coefficient of x
& the given constant value

i.e (+4) * (+8) = +32

Step-2: Now we have
to split the number 32 into two numbers in such a way that after adding them we
get the number equal to coefficient of x i.e 12 & also their multiplication
results into 32.

Therefore we get,
+12= (+4) + (+8)

Also, (+32)= (+4) *
(+8)

Step 3: Now change
the signs of both the factors i.e 4 & 8

After changing the
signs we get, (-4) & (-8)

Step 4: Divide these
factors by the coefficient of x2

Therefore, we get

-4/4=-1

& -8/4= -2

Thus, x= -1, -2

Therefore -1 &
-2 are the required values of x.

Now Let us take one more Example:

Q) 3x2+19x+28=0

Solution:

In the above given equation, we have

+3 is the coefficient of x2

+19 is the coefficient of x

+28 is the constant value

Step-1: Multiply the coefficient
of x & the constant value

i.e (+3) * (+28) =
+84

Step-2: Now we have
to split the number 84 into two numbers in such a way that after adding both
numbers we get the number equal to coefficient of x i.e 19 & also their
multiplication results to 84

Therefore, +19=
(+12) + (+7)

Also, +84 = (+12) *
(+7)

Step 3: Now change the
signs of both the factors

After changing the
signs we get, (-12) & (-7)

Step 4: Now Divide
these factors by the coefficient of x2

Therefore, we get

-12/3=-4

& -7/3= -7/3

Thus, x= -4, -7/3

Therefore -4 &
-7/3 are the required values of x.

How
to Solve More than One Quadratic Equation?

In this type of Quadratic Equation,
more than one quadratic equation will be given. First we have to solve the
given equations individually & find the values of unknown variables given
in the equation & then have to find the relation between these variables of
the equation as given below:

1. x > y

2. x < y

3. x ≥ y

4. x ≤ y

5. x = y or relation
cannot be determined

Example:

x2+13x+40=0 ----(1)

y2+7y+12=0 ----- (2)

First Solve the Equation (1)

Therefore, +13 =
(+5) + (+8)

Also, +40 = (+5) *
(+8)

Now Change the
signs & Divide them with the coefficient of x2 i.e with +1

Therefore, we get

-5/1= -5 &
-8/1= -8

Thus, -5 & -8
are the required values of x.

Now similarly Solve the 2nd Equation,

+7 = (+3) + (+4)

Also +12 = (+3) * (+4)

Now Change the signs & Divide them with the
coefficient of y2 i.e with +1

Remember this diagram friends to solve the Questions of
quadratic equation shortly in the exam but before exam you will need to
practice more and more question to get speed and also to get familiar with the
tricks.

1. I.5x2 – 87x + 378 = 0 II.3y2 – 49y + 200 = 0

I.5x2 – 45x – 42x + 378 = 0

or,5x(x
– 9) – 42(x – 9) = 0

or,(5x –
42) (x – 9) = 0

x = 9,5

II.3y2 – 24y – 25y + 200 = 0

or,3y(y
– 8) –25(y – 8) = 0

or,(y –
8) (3y – 25) = 0

y = 8,3

So , Answers : x > y

Now for this question we can solve this in less
than thirty second how lets see,

Step-1. First take the Equation-I means equation
with variable x.

It is in the form of ax2 + (-)b x + c.

Step-2. See the Sign of b & c means sign of coefficient
of x variable and sign of last one,

It is first negative(-) and second positive(+)

Step -3.From above table we can say that both the
roots of x will be positive.

Step-4.Make Calculation 378*5=1890 first divide by
2 then 3 and then by 5 and follow above rule, so for this it will be 45 and 42 and now
you don't need to write another two to three lines of taking common and then
making in the forms of roots. This is your answer because we know from table
that both root have positive sign.

Step-5 For those question in which coefficient of X2
is not one like in this question you will need to divide both the roots by (a).

So answer will be :-9,5

Follow same step for Equation-II also and find root
in 20-30 seconds but to get this in short time you will need to practice lots
of question then in exam you can solve these questions easily.

Practice Questions :-

1. I.x2 – 8x + 12 = 0 II. y2 – 14y + 49 =
0

2. I.10x2 – x – 24 = 0 II. y2 – 2y + 24 = 0

3. I.x2 – 5x + 6 = 0 II.2y2 – 15y + 27 = 0

4. I.3x + 2y = 301 II.7x – 5y = 74

5. I.14x2 – 37x + 24 = 0 II.28y2 – 53y + 24 = 0

6. I.11x + 5y = 117 II.7x + 13y = 153

7. I.6x2 + 51x + 105 = 0 II.2y2 + 25y + 78 = 0

8. I.6x + 7y = 52 II.14x
+ 4y = 35

9. I.x2 + 11x + 30 = 0 II.y2 + 12y + 36 = 0

10. I.2x2 + x – 1 = 0 II.2y2 – 3y + 1 = 0

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